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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
An important lemma of isogonal conjugate points
buratinogigle   6
N 22 minutes ago by buratinogigle
Source: Own
Let $P$ and $Q$ be two isogonal conjugate with respect to triangle $ABC$. Let $S$ and $T$ be two points lying on the circle $(PBC)$ such that $PS$ and $PT$ are perpendicular and parallel to bisector of $\angle BAC$, respectively. Prove that $QS$ and $QT$ bisect two arcs $BC$ containing $A$ and not containing $A$, respectively, of $(ABC)$.
6 replies
buratinogigle
Mar 23, 2025
buratinogigle
22 minutes ago
A difficult problem [tangent circles in right triangles]
ThAzN1   48
N 30 minutes ago by Autistic_Turk
Source: IMO ShortList 1998, geometry problem 8; Yugoslav TST 1999
Let $ABC$ be a triangle such that $\angle A=90^{\circ }$ and $\angle B<\angle C$. The tangent at $A$ to the circumcircle $\omega$ of triangle $ABC$ meets the line $BC$ at $D$. Let $E$ be the reflection of $A$ in the line $BC$, let $X$ be the foot of the perpendicular from $A$ to $BE$, and let $Y$ be the midpoint of the segment $AX$. Let the line $BY$ intersect the circle $\omega$ again at $Z$.

Prove that the line $BD$ is tangent to the circumcircle of triangle $ADZ$.

comment
48 replies
ThAzN1
Oct 17, 2004
Autistic_Turk
30 minutes ago
IMO 2008, Question 2
delegat   63
N 38 minutes ago by ezpotd
Source: IMO Shortlist 2008, A2
(a) Prove that
\[\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1\] for all real numbers $x$, $y$, $z$, each different from $1$, and satisfying $xyz=1$.

(b) Prove that equality holds above for infinitely many triples of rational numbers $x$, $y$, $z$, each different from $1$, and satisfying $xyz=1$.

Author: Walther Janous, Austria
63 replies
delegat
Jul 16, 2008
ezpotd
38 minutes ago
USAMO 2003 Problem 1
MithsApprentice   69
N an hour ago by de-Kirschbaum
Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd.
69 replies
MithsApprentice
Sep 27, 2005
de-Kirschbaum
an hour ago
Inequalities
sqing   13
N 5 hours ago by ytChen
Let $ a,b>0   $ . Prove that
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1}  \leq  \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1}  \leq  \frac{2}{5} $$
13 replies
sqing
May 13, 2025
ytChen
5 hours ago
System of Equations
P162008   1
N 5 hours ago by alexheinis
If $a,b$ and $c$ are real numbers such that

$(a + b)(b + c) = -1$

$(a - b)^2 + (a^2 - b^2)^2 = 85$

$(b - c)^2 + (b^2 - c^2)^2 = 75$

Compute $(a - c)^2 + (a^2 - c^2)^2.$
1 reply
P162008
Monday at 10:48 AM
alexheinis
5 hours ago
Might be the first equation marathon
steven_zhang123   35
N Yesterday at 7:09 PM by lightsbug
As far as I know, it seems that no one on HSM has organized an equation marathon before. Click to reveal hidden textSo why not give it a try? Click to reveal hidden text Let's start one!
Some basic rules need to be clarified:
$\cdot$ If a problem has not been solved within $5$ days, then others are eligible to post a new probkem.
$\cdot$ Not only simple one-variable equations, but also systems of equations are allowed.
$\cdot$ The difficulty of these equations should be no less than that of typical quadratic one-variable equations. If the problem involves higher degrees or more variables, please ensure that the problem is solvable (i.e., has a definite solution, rather than an approximate one).
$\cdot$ Please indicate the domain of the solution to the equation (e.g., solve in $\mathbb{R}$, solve in $\mathbb{C}$).
Here's an simple yet fun problem, hope you enjoy it :P :
P1
35 replies
steven_zhang123
Jan 20, 2025
lightsbug
Yesterday at 7:09 PM
Maximum number of empty squares
Ecrin_eren   0
Yesterday at 6:35 PM


There are 16 kangaroos on a giant 4×4 chessboard, with exactly one kangaroo on each square. In each round, every kangaroo jumps to a neighboring square (up, down, left, or right — but not diagonally). All kangaroos stay on the board. More than one kangaroo can occupy the same square. What is the maximum number of empty squares that can exist after 100 rounds?



0 replies
Ecrin_eren
Yesterday at 6:35 PM
0 replies
THREE People Meet at the SAME. TIME.
LilKirb   7
N Yesterday at 5:33 PM by hellohi321
Three people arrive at the same place independently, at a random between $8:00$ and $9:00.$ If each person remains there for $20$ minutes, what's the probability that all three people meet each other?

I'm already familiar with the variant where there are only two people, where you Click to reveal hidden text It was an AIME problem from the 90s I believe. However, I don't know how one could visualize this in a Click to reveal hidden text Help on what to do?
7 replies
LilKirb
Monday at 1:06 PM
hellohi321
Yesterday at 5:33 PM
Quite straightforward
steven_zhang123   1
N Yesterday at 3:16 PM by Mathzeus1024
Given that the sequence $\left \{ a_{n} \right \} $ is an arithmetic sequence, $a_{1}=1$, $a_{2}+a_{3}+\dots+a_{10}=144$. Let the general term $b_{n}$ of the sequence $\left \{ b_{n} \right \}$ be $\log_{a}{(1+\frac{1}{a_{n}} )} ( a > 0  \text{and}  a \ne  1)$, and let $S_{n}$ be the sum of the $n$ terms of the sequence $\left \{ b_{n} \right \}$. Compare the size of $S_{n}$ with $\frac{1}{3} \log_{a}{(1+\frac{1}{a_{n}} )} $.
1 reply
steven_zhang123
Jan 11, 2025
Mathzeus1024
Yesterday at 3:16 PM
Inequalities
sqing   0
Yesterday at 2:23 PM
Let $ a,b,c>0. $ Prove that$$a^2+b^2+c^2+abc-k(a+b+c)\geq 3k+2-2(k+1)\sqrt{k+1}$$Where $7\geq k \in N^+.$
$$a^2+b^2+c^2+abc-3(a+b+c)\geq-5$$
0 replies
sqing
Yesterday at 2:23 PM
0 replies
Function and Quadratic equations help help help
Ocean_MathGod   1
N Yesterday at 11:26 AM by Mathzeus1024
Consider this parabola: y = x^2 + (2m + 1)x + m(m - 3) where m is constant and -1 ≤ m ≤ 4. A(-m-1, y1), B(m/2, y2), C(-m, y3) are three different points on the parabola. Now rotate the axis of symmetry of the parabola 90 degrees counterclockwise around the origin O to obtain line a. Draw a line from the vertex P of the parabola perpendicular to line a, meeting at point H.

1) express the vertex of the quadratic equation using an expression with m.
2) If, regardless of the value of m, the parabola and the line y=x−km (where k is a constant) have exactly one point of intersection, find the value of k.

3) (where I'm struggling the most) When 1 < PH ≤ 6, compare the values of y1, y2, and y3.
1 reply
Ocean_MathGod
Aug 26, 2024
Mathzeus1024
Yesterday at 11:26 AM
System of Equations
P162008   1
N Yesterday at 10:30 AM by alexheinis
If $a,b$ and $c$ are complex numbers such that

$\frac{ab}{b + c} + \frac{bc}{c + a} + \frac{ca}{a + b} = -9$

$\frac{ab}{c + a} + \frac{bc}{a + b} + \frac{ca}{b + c} = 10$

Compute $\frac{a}{c + a} + \frac{b}{a + b} + \frac{c}{b + c}.$
1 reply
P162008
Monday at 10:34 AM
alexheinis
Yesterday at 10:30 AM
Inequalities
sqing   19
N Yesterday at 8:40 AM by sqing
Let $ a,b,c>0 , a+b+c +abc=4$. Prove that
$$ \frac {a}{a^2+2}+\frac {b}{b^2+2}+\frac {c}{c^2+2} \leq 1$$Let $ a,b,c>0 , ab+bc+ca+abc=4$. Prove that
$$ \frac {a}{a^2+2}+\frac {b}{b^2+2}+\frac {c}{c^2+2} \leq 1$$
19 replies
sqing
May 15, 2025
sqing
Yesterday at 8:40 AM
Combo problem about regular polygons
NJAX   1
N Apr 27, 2025 by Titibuuu
Source: 1st TASIMO, Day1 Problem3
$Abdulqodir$ cut out $2024$ congruent regular $n-$gons from a sheet of paper and placed these $n-$gons on the table such that some parts of each of these $n-$gons may be covered by others. We say that a vertex of one of the afore-mentioned $n-$gons is $visible$ if it is not in the interior of another $n-$gon that is placed on top of it. For any $n>2$ determine the minimum possible number of visible vertices.

Proposed by David Hrushka, Slovakia
1 reply
NJAX
May 18, 2024
Titibuuu
Apr 27, 2025
Combo problem about regular polygons
G H J
G H BBookmark kLocked kLocked NReply
Source: 1st TASIMO, Day1 Problem3
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NJAX
29 posts
#1 • 2 Y
Y by dmusurmonov, Seungjun_Lee
$Abdulqodir$ cut out $2024$ congruent regular $n-$gons from a sheet of paper and placed these $n-$gons on the table such that some parts of each of these $n-$gons may be covered by others. We say that a vertex of one of the afore-mentioned $n-$gons is $visible$ if it is not in the interior of another $n-$gon that is placed on top of it. For any $n>2$ determine the minimum possible number of visible vertices.

Proposed by David Hrushka, Slovakia
This post has been edited 3 times. Last edited by NJAX, May 18, 2024, 9:41 AM
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Titibuuu
47 posts
#2 • 3 Y
Y by B1t, sami1618, Pomegranat
Suppose first that $n \geq 3$ is even, write $n = 2k$, $k \geq 2$, and consider an $n$-gon $X$ which is not covered by any other (there must clearly be at least one, take any if there are more than one) and an $n$-gon $Y$ which has this property after removing $X$ from the table. Since any diagonal connecting a pair of opposite vertices in $Y$ is the longest among all segments contained in $Y$, and hence also in $X$, no pair of opposite vertices of $Y$ can be covered by the interior of $X$.

It follows that there are at least $k$ visible vertices of $Y$ which together with $2k$ visible vertices from $X$ make a lower bound $3k$.

If $n$ is odd, i.e., $n = 2k + 1$ for some $k \geq 1$, we choose $X$ and $Y$ in the same way and moreover pick $Z$ being ``on top'' after both $X$ and $Y$ are removed. Then we argue analogously as above (using the fact that the diagonal connecting a vertex $V$ of an $(2k+1)$-gon $G$ with any endpoint of the opposite side to $V$ in $G$ realizes the longest segment contained in $G$) to conclude that at least $k + 1$ vertices from $Y$ are visible and at least one vertex of $Z$ must be visible. Thus, we get a lower bound of
\[
(2k+1) + (k+1) + 1 = 3k+3.
\]
The following constructions prove that those lower bounds are achievable for any $n \geq 3$.

First, consider the case $n = 2k$ again. Take three copies $P, Q, R$ of a regular $2k$-gon, place $R$ with its side $AB$ going horizontally and being the bottom one, $Q$ precisely on top of $R$, and $P$ precisely on top of $Q$. Then rotate $Q$ around $A$ and $P$ around $B$, one clockwise, the other counterclockwise, both by equally large angles so that the segment $AB$ is not covered by $P$ and $Q$ anymore and the opposite vertex of $A$ in $Q$ and the opposite vertex of $B$ in $P$ meet.

It is easy to see that only $A$ and $B$ are now visible among the vertices of $R$. Shifting finally $P$ and $Q$ down by distances $d$ and $2d$ for some $d>0$ small enough makes no vertices of $R$ visible and precisely $k$ vertices of $Q$ visible. The remaining $2019$ $2k$-gons can be placed identically as $R$ (under it) which makes in total $2k+k=3k$ visible vertices.

Secondly, let $n = 2k+1$. Analogously, place four copies of a $(2k+1)$-gon on the table, one precisely on top of the previous ones, one common side lying horizontally and below. Then observe that shifting the top layer to the right, the second one to the left, then the first one down, then the second one also down (but less), and finally the third one up, with all distances chosen properly (and after hiding $2018$ more copies of our $(2k+1)$-gon under the fourth layer) we get $(2k+1) + (k+1) + 1 = 3k+3$ visible vertices.
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